# Prove the following equation

• Oct 25th 2012, 11:22 AM
Ogniok
Prove the following equation
Hi. How to prove that equation:

$|AB|^{2} = |QA| * |AE| + |NB| * |BD|$

$|BN| = |BC|$
$|AQ| = |AC|$

$|AD|$ and $|BD|$ are heights of acute-angled triangle $ABC$.

Illustration: http://i.imgur.com/9O8Bm.jpg

I would really appreciate any help. http://www.mymathforum.com/images/sm...on_biggrin.gif
• Oct 25th 2012, 03:05 PM
BobP
Re: Prove the following equation
First point is that most of the diagram is unnecessary.
If $AQ=AC$ you don't need $Q$ or $P.$
If $BN=BC$ you don't need $N$ or $M.$
Further, $L$ and $K$ serve no purpose.
All that you need is the triangle $ABC$ and the perpendiculars $AD$ and $BE.$

Using Pythagoras twice, with the triangles $ADB$ and $ADC,$

$AB^{2}=AD^{2}+BD^{2}=(AC^{2}-CD^{2})+BD^{2}=AC^{2}-(BC-BD)^{2}+BD^{2}=AC^{2}-(BC^{2}+BD^{2}-2BC.BD)+BD^{2}=AC^{2}-BC^{2}+2BC.BD.$

Now repeat that sequence using the triangles $ABE$ and $BEC.$

Having done that, add the resulting equation to the one that's just been derived.
• Oct 26th 2012, 07:29 AM
Ogniok
Re: Prove the following equation
Thank you very much. :D I've finally done that :D

Here are full calculations:

$AB^{2} = AD^{2} + BD^{2} = (AC^{2} - CD^{2}) + BD^{2} = AC^{2} - (BC - BD)^{2} + BD^{2} = AC^{2} - (BC^{2} + BD^{2} - 2 * BC * BD) + BD^{2} = AC^{2} - BC^{2} + 2 * BC * BD$

$AB^{2} = BE^{2} + AE^{2} = (BC^{2} - CE^{2}) + AE^{2} = BC^{2} - (AC - AE)^{2} + AE^{2} = BC^{2} - (AC^{2} + AE^{2} - 2 * AC * AE) + AE^{2} = BC^{2} - AC^{2} + 2 * AC * AE$

$2AB^{2} = AC^{2} - BC^{2} + 2 * BC * BD + BC^{2} - AC^{2} + 2 * AC * AE$
$2AB^{2} = 2 * BC * BD + 2 * AC * AE$
$AB^{2} = BC * BD + AC * AE$
• Nov 1st 2012, 03:02 AM
elisaevedent
Re: Prove the following equation